Prove with induction: $\forall n \in \mathbb{N}_{>2} \: \exists y\in \mathbb{N}: (n+2)^3 + 2(n+2) = 3y$

I think @user21820HATESSMOKING-HATS exaggerates how much you need to formally understand, so let's work with as few symbols as possible. I want you to take this as a guide to how a proof by (weak) induction is supposed to be formatted; that, rather than solving this specific example, is its focus.

Theorem: for integer $m\ge5$, some integer $y$ satisfies $m^3+2m=3y$. (My $m$ is your $n+2$, but using $m$ will tidy what we do next.)

Base case of proof by induction: I'll leave you to calculate which $y\in\Bbb Z$ works.

Inductive step of proof by induction: check if the case $m=k$ works so does the case $m=k+1$, i.e. if $k^3+2k=3y$ then $(k+1)^3+2(k+1)=3(y+z)$ for some integer $z$ (work through the algebra yourself to work out which integer). Your stab at the problem didn't verify the increase is thrice an integer; it just wrote an integer as thrice a fraction.

At this point a pedant will argue using a base case $m=5$ isn't what the axiom schema of induction covers, as it's written to use $m=0$. But with $t:=m-5$, we could have stated all this in terms of what's true for integer $t\ge0$; $m$ as I've defined it just makes the algebra simpler for pedagogy. (The "pedant" in this case would know all that, but would want the proof-writer to spell it out properly.) Therefore, I invite you to do a more complicated version starting at $0$.

Of course, we can show $3|m^3+2m$ for all integer $m\ge0$ (or even for arbitrary $m\in\Bbb Z$, but that's another story), so the fact $m=n+2$ appeared in your original problem statement makes me think $\forall n\in\Bbb N_{>2}$ should have been $\forall n\in\Bbb Z_{\ge-2}$. Then the base $m$ would have actually been $0$.